Kinetic Theory

JEE Main level, single correct, medium

47 of 168

Question 47 of 168

If the R.M.S. speed of oxygen molecules at 0∘C0^{\circ}C is 160ms−1160m{s}^{-1}. Find the R.M.S. speed of hydrogen molecules at 0∘C0^{\circ}C.

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Answer: (C) 640ms−1640m{s}^{-1}

Recall the formula of root mean square speed, vRMS=3RTM{v}_{RMS}=\sqrt{\dfrac{3RT}{M}},
⇒(vRMS)O2(vRMS)H2=MH2MO2=232=14\Rightarrow \dfrac{{\left({v}_{RMS}\right)}_{{O}_{2}}}{{\left({v}_{RMS}\right)}_{{H}_{2}}}=\sqrt{\dfrac{{M}_{{H}_{2}}}{{M}_{{O}_{2}}}}=\sqrt{\dfrac{2}{32}}=\dfrac{1}{4}
⇒(vRMS)H2=4(vRMS)O2=4×160=640m s−1\Rightarrow {\left({v}_{RMS}\right)}_{{H}_{2}}=4{\left({v}_{RMS}\right)}_{{O}_{2}}=4\times 160=640{\text{m s}}^{-1}

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